Tag: counting problems

Posted in Combinatorics, High school mathematics

Linking combination and permutation with repetition

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Combinatorial problems are difficult because it’s hard to  know which formula to use in a particular problem and when you need to ‘tweak’ or totally abandon the formula. In this post I share two solutions to a problem which connects the multiplication  principle, the combination formula and the formula for counting the number of permutation with repetition. Knowledge of connections among concepts help in problem solving. The problem is generated from the rook puzzle presented in my post Connecting Pascal’s Triangle and Permutation with identical objects.

Find the number of different ways of arranging 14 letters 7 of which are E’s and 7 are N’s, in a row. Here is one arrangement:

The first solution shows how the formula for counting permutation with identical objects can be deduced from the solution involving the multiplication principle and the second connects permutation and combination.

Solution 1

The idea behind this solution is to initially treat each letter as distinct. There are 14 letters to be arranged in a row. If these letters are distinct from one another then by the Multiplication Principle, there are 14! different arrangements in all.

But the letters are not all distinct. In fact there are only two kinds – N’s and E’s. This means an arrangement, for example, consisting of 7 N’s followed by 7 E’s,

N E N E N E N E N E N E N E 

has been counted 7!7! times in all in 14!. The same is true for this arrangement:

N N N E E N N E E N N E E E.

Thus, to find the number of ways of arranging 14 letters where 7 of which are identical and the remaining 7 are also identical, 14! need to be divided by 7!7!

If the problem had been In how many ways can you arrange 10 N‘s and 4 E‘s?, the solution will be \frac {14!}{10!4!}.

Notice that this solution uses the technique of counting the number of permutations (arrangements) of n objects, r1 of which are identical, r2 are identical, . . . , and rn are identical, where r1 + r2 + . . . ri = n. The number of different permutations is denoted by

In the problem, n = 14, r1 = 7 and r2 = 7. Hence, the total number of arrangements is  \frac{14!}{7!7!} = 3432.

Solution 2

This solution simplifies the original problem to How many different ways can 7 N’s be arranged in a row of 14 spaces? Now, why is this problem equivalent to the original problem? What happened to the 7 E‘s? Why aren’t they not considered anymore? This is because for a particular arrangement of 7 N‘s in 14 spaces, there is one and only one way the 7E‘s can be arranged.

To solve, count the number of possible positions for the 7 N’s. You have 14 positions to choose from for the first N. For the next N you only have 13 positions to choose from, for the next N, 12 and so on until the 7th N. By the Multiplication Principle you have  14.13.12.11.10.9.8.7 different possible positions where you consider each N to be distinct.

But the N’s are identical. That is, in the arrangement for example

_ N _ N N N _ N N _ _ _ _ N

N has been counted 7! times in 14.13.12.11.10.9.8.7. So you have to divide this by 7!

Thus, there are  different positions for N (all N are identical). This can be written in as shorter way using factorial notation by multiplying it by \frac{7!}{7!}.

If there were 10N‘s and 4 E‘s, the problem would have been In how many ways can I arrange 10 N’s in 14 spaces?

In general, if there were n possible positions for arranging r objects, the formula is \frac {n!}{r!(n-r)}. Note that this looks like the combination formula which is used to solve problems for the number of combination and indeed it is. You just got used to applying nCr = \frac {n!}{r!(n-r)} to problems like In how many way can you arrange n different objects taken r objects at a time?

Posted in Combinatorics

What makes counting problems difficult

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I don’t have the numbers but I think not many will disagree with me that among those who like mathematics and find joy in the challenge of solving math problems, only very few will also like to solve combinatorial or counting problems. And if I may be allowed to push my observation a bit further, those who are more inclined to algebra are the ones who have more aversion to problems that asks them to count.

Indeed, combinatorial problems are odd. Unlike other mathematics problems, these type of problems cannot easily be categorized and solved with predictable algorithms. Each problem always seems to be a case in itself. Knowing all the formulas for different cases of permutations and combinations is not a guarantee that one will be able to solve these problems. For many, these are only useful if the problem already explicitly asks for number of permutations and combinations. And when indeed the problem says so, one still has to decipher other conditions and the assumptions embedded in the context. Knowing when to add or multiply is also difficult, if not the most difficult part. Knowing when one already double counted also requires depth of understanding not only of the formulas but also of the context of the problem. Of course for some these are also the very reason why they love to solve counting problems.

Take the case of this family of problems:

 

  1. How many three-digit numbers can you form from the digits, 5, 6, 7, 8, and 9? (This requires direct application of the permutation formula or better, the multiplication principle.)
  2. How many three-digit numbers are there in all? (That you will choose from 0 to 9 is implied only plus you ask yourself how about 027? Is it a 3-digit number of a 2-digit number?)
  3. How many numbers can be formed from 5, 6, 7, 8, and 9?
  4. How many three-digit numbers greater that 600 can be formed from the digits 5, 6, 7, 8, and 9? (Oops … so what formula is applicable now?)
  5. How many numbers greater than 3,000,000 can be formed by arrangements of the digits 1, 2, 2, 3, 6, 6, 6? (More Ooops…)
Having identified some of the difficulties inherent to counting problems, what teaching tip can I propose?
  1. Ask for different ways of solving the problem, slowly discouraging students’ reliance to listing. Thinking in terms of tree diagrams and empty boxes are great help.
  2. Defer introduction of formulas before students can solve the problems like the ones above using the fundamental counting principle. Permutations and combination formulas are only powerful to the extent to which students understand the multiplication principle.
  3. Another pedagogical tip is to ask students to identify what makes a problem similar and different in structure from the problems they previously solved. If they can specify which problem, the better.
  4. As a consequence of tip#3, problems are best given per family with family defined in terms context (e.g., the family of problems given in the example above) and not in terms of similarity in solution structure.

Please add yours. Thanks.

You may want to read my two other posts on combinatorics:

Some useful references: